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To find the voltage \( V \) of the circuit, we use Ohm's Law, which is given by:

\[
V = I \times R
\]

Where:
- \( I = 2 - 3i \) amperes (current)
- \( R = 3 + 5i \) ohms (resistance)

We need to multiply the two complex numbers \( I \) and \( R \) to find the voltage \( V \).

\[
V = (2 - 3i) \times (3 + 5i)
\]

Now, we can expand this using the distributive property:

\[
V = (2)(3) + (2)(5i) + (-3i)(3) + (-3i)(5i)
\]

Simplifying each term:

\[
V = 6 + 10i - 9i - 15i^2
\]

Since \( i^2 = -1 \), we have:

\[
V = 6 + 10i - 9i + 15
\]

Now, combine like terms:

\[
V = (6 + 15) + (10i - 9i) = 21 + i
\]

Thus, the voltage \( V \) is:

\[
V = 21 + i \, \text{volts}
\]

Would you like more details on any part of this explanation or have any questions?

Here are 5 questions to explore further:
1. How do complex numbers represent quantities in AC circuits?
2. Can Ohm's Law be applied in circuits involving purely real values?
3. How do you convert complex impedance to polar form?
4. What is the physical significance of the imaginary part of impedance?
5. What happens when current or resistance are purely imaginary?

**Tip:** Always remember to combine like terms carefully when multiplying complex numbers, especially handling \( i^2 = -1 \).

Find V, the voltage of the circuit, if I = (2 - 3i) amperes and R = (3 + 5i) Ohms.

Learn how to find the voltage in an electrical circuit involving complex current and resistance using Ohm's Law. This problem covers key concepts such as complex number multiplication and is suitable for students in grades 11-12 or early university-level circuits.

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